= Solution
Set $h=x+3$, so $x=-3(1-h/3)$. The <negative binomial series> gives $(1-u)^{-4}=\sum_{n=0}^\infty\binom{n+3}{3}u^n$ for $|u|<1$. Hence the exact <Taylor series> is
$$
\boxed{\frac7{x^4}=\frac7{81}\sum_{n=0}^\infty\binom{n+3}{3}\left(\frac{x+3}{3}\right)^n,\qquad|x+3|<3.}
$$
The initial terms are $\frac7{81}[1+4h/3+10h^2/9+20h^3/27+\cdots]$. Equivalently the coefficient of $(x+3)^n$ is $7(n+3)!/[6n!3^{n+4}]$, obtained by differentiating $7x^{-4}$ repeatedly and evaluating at $-3$.
The radius of convergence is three, set by the pole at $x=0$. At either endpoint $h=\pm3$, the terms fail to tend to zero, so neither endpoint is included.
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